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Ch 13: Newton's Theory of Gravity
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 13, Problem 52b

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.0 x 1016 kg and a radius of 8.8 km. What is the escape speed from the asteroid?

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Step 1: Recall the formula for escape speed, which is derived from the conservation of energy principle. The escape speed \( v_{\text{esc}} \) is given by \( v_{\text{esc}} = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \), \( M \) is the mass of the asteroid, and \( R \) is the radius of the asteroid.
Step 2: Convert the radius of the asteroid from kilometers to meters. Since \( 1 \, \text{km} = 1000 \, \text{m} \), the radius \( R \) becomes \( 8.8 \times 10^3 \, \text{m} \).
Step 3: Substitute the given values into the escape speed formula. Use \( M = 1.0 \times 10^{16} \, \text{kg} \), \( R = 8.8 \times 10^3 \, \text{m} \), and \( G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \). The formula becomes \( v_{\text{esc}} = \sqrt{\frac{2 \cdot (6.674 \times 10^{-11}) \cdot (1.0 \times 10^{16})}{8.8 \times 10^3}} \).
Step 4: Simplify the numerator \( 2 \cdot G \cdot M \) and the denominator \( R \) separately. This will help in organizing the calculation process.
Step 5: Take the square root of the resulting value from the division \( \frac{2GM}{R} \) to find the escape speed \( v_{\text{esc}} \). Ensure units are consistent throughout the calculation, and the final speed will be in meters per second (\( \text{m/s} \)).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Escape Velocity

Escape velocity is the minimum speed an object must reach to break free from the gravitational pull of a celestial body without any additional propulsion. It depends on the mass of the body and the distance from its center to the point of escape, typically calculated using the formula v = √(2GM/r), where G is the gravitational constant, M is the mass of the body, and r is its radius.
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Gravitational Constant

The gravitational constant (G) is a fundamental physical constant used in the calculation of gravitational forces and escape velocity. Its value is approximately 6.674 × 10⁻¹¹ N(m/kg)². This constant plays a crucial role in determining the strength of the gravitational attraction between two masses, influencing how quickly an object can escape a celestial body's gravity.
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Mass and Radius of Celestial Bodies

The mass and radius of a celestial body, such as an asteroid, are critical parameters in calculating escape velocity. The mass determines the strength of the gravitational field, while the radius affects the distance from the center of mass to the surface. In this case, the asteroid's mass is 1.0 x 10¹⁶ kg and its radius is 8.8 km, both of which are essential for applying the escape velocity formula.
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Textbook Question

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Textbook Question

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